`int 1/((x+a)(x+b)) dx` Evaluate the integral

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Chapter 7, 7.4 - Problem 14 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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kspcr111 | In Training Educator

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`int 1/((x+a)(x+b)) dx`


`int 1/((x+a)(x+b)) dx`

let `[u= x+a] => du = dx`

so ,

 `int 1/((x+a)(x+b)) dx`

=`int 1/((u)(x+b)) du`

As `u =x+a `  => `x= u-a` , on substituting in the above equation  we get ,

=`int 1/((u)(x+b)) du`

=`int 1/((u)((u-a)+b)) du`

=`int 1/((u)(u+b-a)) du`

Taking partial fractions we obtain,

=`int 1/((u)(u+b-a)) du`

 `=int ((1/((a-b)(u+b-a)))+ (1/(u*(b-a)))) du`

=`int (1/((a-b)(u+b-a)))du+ int (1/(u*(b-a))) du` -----------------(1)

Now let us consider 

`int (1/((a-b)(u+b-a)))du`

let `v=u+b-a `  => `dv =du`


`int (1/((a-b)(u+b-a)))du`

= `(1/(a-b)) int (1/(u+b-a))du`

=`(1/(a-b)) int (1/(v))dv`

=`(1/(a-b)) ln(v)`

=`(1/(a-b)) ln(u+b-a)`

=`(1/(a-b)) ln(x+b) `  as `[u-a =x] `  ----------------------(2)

Now consider ,

`int (1/(u*(b-a))) du`

As similar to above we obtain as follows,

`int (1/(u*(b-a))) du`

= `(1/(b-a))int (1/(u)) du`

=`(1/(b-a)) ln(u)`

= `(1/(b-a)) ln(x+a)`  as `u=x+a` -------------(3)

substituting (2) and (3) in (1) we get,

`int 1/((x+a)(x+b)) dx`

=`int (1/((a-b)(u+b-a)))du+ int (1/(u*(b-a))) du`

=    `(1/(a-b)) ln(x+b) +(1/(b-a)) ln(x+a) + C`

is the solution :)

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